# Finding all the solutions to $\sin(5x) - \sin (3x) = \sqrt 2 \; \cos(4x)$

$$\sin(5x) - \sin (3x) = \sqrt 2 \; \cos(4x)$$

After working with the equation, I got
$$2\sin(x)\cos(4x) = \sqrt 2 \; \cos(4x)$$ with difference of sines formula.

I saw that, I need to check if $$\cos(4x)=0$$ is a solution, so I got $$x= \frac{\pi}{8}$$ and it worked, so my first two solutions are

$$x= \frac{\pi}{8} + 2\pi n \space, n \in \Bbb Z \quad\text{and}\quad x= -\frac{\pi}{8} + 2\pi n \space, n \in \Bbb Z$$

After that, I divided by $$\cos(4x)$$, getting $$\sin(x) = \frac{\sqrt 2}{2}$$, so

$$x= \frac{\pi}{4} + 2\pi n \space, n \in \Bbb Z \quad\text{and}\quad x= \frac{3\pi}{4} + 2\pi n \space, n \in \Bbb Z$$

are another 2 solutions.

I checked in Wolfram Alpha to see if my work is correct, and I found that there are $$10$$ solutions to this equation. How can I get the other $$6$$ solutions I'm missing? (My 4 solutions are correct).

Solution to $$\cos \alpha = 0$$ is $$\alpha = {\pi \over 2} +\pi n$$
so in your case $$x ={\pi \over 8} +{\pi \over 4}n$$
$$n\in\mathbb{Z}$$.
From $$\cos 4x=0$$ one gets $$4x=\pm\frac\pi 2+2n\pi$$, that is $$x=\pm\frac{\pi}8+\frac{n\pi}2$$. That's eight solutions modulo $$2\pi$$.
From $$\sin x=\frac1{\sqrt2}$$ one gets $$x=\frac\pi4+2n\pi$$ and $$x=\frac{3\pi}4+2n\pi$$, that two more solutions modulo $$2\pi$$.
• Why the person that answered before you put one of the solution as $x ={\pi \over 8} +{\pi \over 4}n$ instead of $x ={\pi \over 8} +{\pi \over 2}n$ ? Commented Jul 13, 2019 at 6:08
• @RodrigoPizarro : Because $-\frac\pi+(2n)\pi=\frac\pi2+(2n-1)\pi$, so that the set of $\pm\frac\pi2+(2n)\pi$ is the same as the set of $\frac\pi2+k\pi$. Commented Jul 13, 2019 at 7:23